Let A=${a_n A=$\{a_n : n\in \omega }\subset \}\subset 2^{\omega\times\omega}$be nonempty countable without isolate point(i.e Homeomorphism isolated points (i.e. homeomorphic to Q),$\mathbb{Q}$), and satisfy$ \forall n\in \omega \exists^\infty m|{k:a_n(m,k)=1}|=\omega m|\{k:a_n(m,k)=1\}|=\omega $. Is Does there exist$ a\in cl(A)\setminus A$satisfy satisfying$\exists^\infty m|{k:a(m,k)=1}|=\omega m|\{k:a(m,k)=1\}|=\omega $? 4 added 3 characters in body; edited body; edited body Let$A={a_n:n\in A=${a_n : n\in \omega }\subset 2^{\omega\times\omega}$ be nonempty countable without isolate point(i.e Homeomorphism to Q), and satisfy $\forall n\in \omega \exists^\infty m|{k:a_n(m,k)=1}|=\omega$. Is there exist $a\in cl(A)\setminus A$ satisfy $\exists^\infty m|{k:a(m,k)=1}|=\omega$?
Let $A=(a_n:n\in A={a_n:n\in \omega )}\subset 2^{\omega\times\omega}$ be nonempty countable without isolate point(i.e Homeomorphism to Q), and satisfy $\forall n\in \omega \exists^\infty m|(k:a_n(m,k)=1)|=\omega m|{k:a_n(m,k)=1}|=\omega$. Is there exist $a\in cl(A)\setminus A$ satisfy $\exists^\infty m|(k:a(m,k)=1)|=\omega m|{k:a(m,k)=1}|=\omega$?